If the second term in the expansiona'13+aa-1nis 14a52 then the value of C3 nC2 n is
We have, T2=14a52 ⇒C1 na113n-1a321=14a52 ⇒nan-113+32=14a52 ⇒n=14 ⇒C3 nC2 n=C3 14C2 14=123=4